cos alpha sin (beta-gamma) + cos beta sin (ga | vedclass

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Question

(A) Expand the expression using the formula $\sin (A-B) = \sin A \cos B - \cos A \sin B$: $\cos \alpha (\sin \beta \cos \gamma - \cos \beta \sin \gamm

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(A) Expand the expression using the formula $\sin (A-B) = \sin A \cos B - \cos A \sin B$: $\cos \alpha (\sin \beta \cos \gamma - \cos \beta \sin \gamma) + \cos \beta (\sin \gamma \cos \alpha - \cos \gamma \sin \alpha) + \cos \gamma (\sin \alpha \cos \beta - \cos \alpha \sin \beta)$ Distribute the terms: $\cos \alpha \sin \beta \cos \gamma - \cos \alpha \cos \beta \sin \gamma + \cos \beta \sin \gamma \cos \alpha - \cos \beta \cos \gamma \sin \alpha + \cos \gamma \sin \alpha \cos \beta - \cos \gamma \cos \alpha \sin \beta$ Grouping the terms,we see that each term cancels out: $(\cos \alpha \sin \beta \cos \gamma - \cos \gamma \cos \alpha \sin \beta) + (-\cos \alpha \cos \beta \sin \gamma + \cos \beta \sin \gamma \cos \alpha) + (-\cos \beta \cos \gamma \sin \alpha + \cos \gamma \sin \alpha \cos \beta) = 0 + 0 + 0 = 0$

$\cos \alpha \sin (\beta-\gamma) + \cos \beta \sin (\gamma-\alpha) + \cos \gamma \sin (\alpha-\beta)$ is equal to

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